A particle is moving with speed `6m//s` along the direction of `vec(A)=2hat(i)+2hat(j)-hat(k)`, then its velocity is `:`

Options

A. `(4hat(i)+2hat(j)-4hat(k))` units
B. `(4hat(i)+4hat(j)-2hat(k))` units
C. `(4hat(i)+4hat(j)-4hat(k))` units
D. `(2hat(i)+4hat(j)-2hat(k))` units

Explanation

To find the velocity vector of a particle moving with a speed of \(6 \, \text{m/s}\) along the direction of the vector \(\vec{A} = 2\hat{i} + 2\hat{j} - \hat{k}\), we can follow these steps:

### Step 1: Identify the direction vector
The direction vector is given as:
\[ \vec{A} = 2\hat{i} + 2\hat{j} - \hat{k} \]

### Step 2: Calculate the magnitude of the direction vector
The magnitude of vector \(\vec{A}\) is calculated using the formula:
\[ |\vec{A}| = \sqrt{(2)^2 + (2)^2 + (-1)^2} \]
Calculating this:
\[ |\vec{A}| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 3: Find the unit vector in the direction of \(\vec{A}\)
The unit vector \(\hat{a}\) in the direction of \(\vec{A}\) is given by:
\[ \hat{a} = \frac{\vec{A}}{|\vec{A}|} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3} \]

### Step 4: Calculate the velocity vector
The velocity vector \(\vec{v}\) can be found by multiplying the unit vector \(\hat{a}\) by the speed of the particle:
\[ \vec{v} = |\text{speed}| \cdot \hat{a} = 6 \cdot \hat{a} = 6 \cdot \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3} \]
This simplifies to:
\[ \vec{v} = 2(2\hat{i} + 2\hat{j} - \hat{k}) = 4\hat{i} + 4\hat{j} - 2\hat{k} \]

### Final Answer
Thus, the velocity vector is:
\[ \vec{v} = 4\hat{i} + 4\hat{j} - 2\hat{k} \]

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